The differential rate law for the reaction $H_2 + I_2 \to 2HI$ is:
- A$-\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = +\frac{1}{2}\frac{d[HI]}{dt}$
- B$-\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = +\frac{d[HI]}{dt}$
- C$\frac{1}{2}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[I_2]}{dt} = -\frac{d[HI]}{dt}$
- D$-2\frac{d[H_2]}{dt} = -2\frac{d[I_2]}{dt} = +\frac{d[HI]}{dt}$
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Similar Questions
What is the rate of formation of $O_2$ for the reaction stated below?
$2 N_2O_{5(g)} \longrightarrow 4 NO_{2(g)} + O_{2(g)}$
Given: $\left| \frac{d[N_2O_5]}{dt} \right| = 0.02 \ mol \ dm^{-3} \ s^{-1}$
$2 N_2O_{5(g)} \longrightarrow 4 NO_{2(g)} + O_{2(g)}$
Given: $\left| \frac{d[N_2O_5]}{dt} \right| = 0.02 \ mol \ dm^{-3} \ s^{-1}$
For the reaction $2A + B \rightarrow 3C + D$,which of the following does not express the rate of reaction?
Medium
View SolutionConsider the following reaction: $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$. The rate of this reaction in terms of $N_2$ at $T \ K$ is $\frac{-d[N_2]}{dt} = 0.02 \ mol \ L^{-1} \ s^{-1}$. What is the value of $\frac{-d[H_2]}{dt}$ (in units of $mol \ L^{-1} \ s^{-1}$) at the same temperature?
In the reaction $2SO_3 \to 2SO_2 + O_2$,initially $40 \ g$ of $SO_3$ is heated in a $100 \ mL$ container. If after $10 \ min$,$3.2 \ g$ of $O_2$ is formed,find the rate of reaction in $mol \ L^{-1} \ min^{-1}$.
Medium
View SolutionFor the reaction $N_{2}O_{5(g)} \rightarrow 2NO_{2(g)} + \frac{1}{2} O_{2(g)}$,the rate of disappearance of $N_{2}O_{5}$ is given as $6.25 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$. The rate of formation of $NO_{2}$ and $O_{2}$ is given respectively as:
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